If $k$ is a commutative field of characteristic $p>0$, then the map $$ \theta \colon \mathrm{GL}_p(k) \to \mathrm{SL}_p(k) \colon A = (a_{ij}) \mapsto (\det A)^{-1} (a_{ij}^p) $$ is a group homomorphism. Does this have a name? It is, of course, very closely related to the Frobenius endomorphism on $\mathrm{GL}_p(k)$ but it is not quite the same.
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asked Jun 23, 2023 at 13:59
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$\begingroup$ @GeoffRobinson I'm not assuming $k$ to be perfect (and in fact, I also want to consider this for $k$ replaced by a $k$-algebra $R$, in the context of algebraic groups). $\endgroup$– Tom De MedtsCommented Jun 23, 2023 at 15:47
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$\begingroup$ @GeoffRobinson Sorry for insisting, but I think that when $k$ is not perfect, there is only a non-canonical way to view $\mathrm{GL}_p(k)$ as such a direct product, via the natural inclusion of $\mathrm{SL}_p(k)$ together with the subgroup of, e.g., diagonal matrices $\operatorname{diag}(a, 1, \dots, 1)$. So in that sense, it seems to me that my map $\theta$ is quite different from the projection map. (Sorry if I am misunderstanding what you are trying to say.) $\endgroup$– Tom De MedtsCommented Jun 24, 2023 at 11:56
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1$\begingroup$ Yes, you are right, and I am wrong (again!). I was thinking that the scalar matrices form a complement to ${\rm SL}(p,k)$, but I was stuck in the world of finite groups. $\endgroup$– Geoff RobinsonCommented Jun 24, 2023 at 12:08
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3$\begingroup$ The map $\theta$ is not really a projection since it's not idempotent. It's the Frobenius map twisted by a character, but indeed this very one has the special property of having image in $\mathrm{SL}_n$). Since it defines a certain special $p$-dimensional representation of $\mathrm{GL}_n$, maybe the RT tag would be useful. $\endgroup$– YCorCommented Jun 24, 2023 at 17:33
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